[Leetcode] Remove Nth Node From End of List

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本文介绍了一种高效算法，用于删除单链表中倒数第N个节点，仅通过一次遍历实现。该算法适用于1->2->3->4->5这样的链表结构，例如当N为2时，操作后的链表变为1->2->3->5。

Given a linked list, remove the n^th node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

class Solution{
public: 
	ListNode *removeNthFromEnd(ListNode *head, int n) {
		if(!head) return NULL;
		ListNode* current = head;
		int count = 0;
		if(n==0) return head;
		while(current!=NULL)
		{
			current = current->next;
			count++;
		}

		//if(n==count) return head->next;
		int index = count - n;
		count = 0;

		ListNode* vHead = new ListNode(0);
		vHead->next = head;
		current = vHead;
		while(count<index)
		{
			current = current->next;
			count++;
		}
		ListNode *tmp = current->next->next;
		current->next = tmp;

		return vHead->next;

	}
}